$d(n) = 3 \left(-2\right)^{n - 1}$ What is the $4^\text{th}$ term in the sequence?
This is an explicit formula. All we have to do is plug $n=4$ in the formula to find the $4^\text{th}$ term. $\begin{aligned} d({4}) &=3(-2)^{{4} - 1} \\\\ &= -24 \end{aligned}$ The $4^\text{th}$ term is $-24$.